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User:Neocapitalist

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| style="font-size: 8pt; padding: 4pt; color: white; line-height: 1.25em;" | This user is a Capitalist.

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That which is not forbidden is compulsory.

This is the law of human existence.

Math

=other stuff=

\mathbf{v} \in \mathbb{R}^n

\mathbf{v} = \langle v_1, v_2, v_3, ..., v_n \rangle

=points on a sphere=

The volume of a sphere is given by

\frac{4\pi}{3}R^3;

Then

\frac{dV}{dt} = \frac{4\pi}{3}3r^2\frac{dR}{dt} = 4\pi r^2\frac{dr}{dt}.

Since the infinitesimal distance on the surface of the sphere is given by

d\theta\ ^2 + d\phi\ ^2 = dr^2,

differentiating gives

\frac{d\theta}{dt} d\theta + \frac{d\phi}{dt} d\phi = \frac{dr}{dt} dr = \left (\frac{1}{4\pi r^2} \right) \frac{dV}{dt} dr

=points on a hypersphere=

The volume of a hypercube should be given by

\int_{-R}^{R} \left (\int_{-y}^{y} \left (\sqrt{R^2-x^2-y^2}\right )^3 dx \right) dy

Then, if I have my numbers right,

=Proof there exists a number of the form <math>a^n</math> s.t. a, n are irrational, and <math>a^n</math> is rational=

Let k=n=\sqrt{2}. Then k^n=\sqrt{2}^\sqrt{2}. Now, \sqrt{2}^\sqrt{2} is either rational or irrational. If it is rational, then the theorem is proven. If it is irrational, then we can choose a new a=(\sqrt{2}^\sqrt{2}). Then (\sqrt{2}^\sqrt{2})^\sqrt{2}=\sqrt{2}^2=2, which is rational.

=Riemann=

\zeta(z)=\sum_{n=0}^\infty \frac{1}{n^z} \zeta(z)=0\wedge z\ne-2n\Rightarrow Re(\zeta(z))=\frac{1}{2}

=Wiggy=

\int \equiv